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Proof step 5 from proof in-del-01 (previous step)

r -- sk ∈ st -- r0, ∀ r1, sk0. r0 ≠ r1 → (r1 -- sk0 ∈ st ↔ r1 -- sk0 ∈ st -- r0) ⊦ r ≠ r0 ∧ r -- sk ∈ st, r0 ∈ st -- r0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ r -- sk ∈ st → r ∈ st
Proof step 4:
r -- sk ∈ st -- r0, r ∈ st -- r0, ∀ r1, sk0. r0 ≠ r1 → (r1 -- sk0 ∈ st ↔ r1 -- sk0 ∈ st -- r0) ⊦ r = r0, r0 ∈ st -- r0, r -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]